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Q.What is an electric dipole? Derive an expression for the electric field intensity at a point on the axis of the dipole. (1+6=7)

Odisha ChseOdisha CHSE +2 Science Board Exam 2018Subjective· 7mImportance★★★★★
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Dipole = +q and -q separated by 2a, p = q(2a). Axial field E = (1/4 pi epsilon0) 2pr/(r^2 - a^2)^2, and for r >> a, E = 2kp/r^3 along p.

Definition: An electric dipole is a system of two equal and opposite point charges, +q+q and −q-q, separated by a small distance 2a2a. Its dipole moment is a vector p⃗\vec p of magnitude p=q(2a)p = q(2a) directed from the negative to the positive charge.

Axial field (point P on the axis, at distance rr from the centre O):

Step 1 — Distances from P to the two charges:

to +q+q: (r−a)(r - a); to −q-q: (r+a)(r + a).

Step 2 — Field due to +q+q (pointing away, along the axis outward):

E+=14πε0q(r−a)2E_+ = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{(r-a)^2} (directed from +q+q toward P).

Field due to −q-q (pointing toward −q-q):

E−=14πε0q(r+a)2E_- = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{(r+a)^2} (directed toward −q-q, opposite to E+E_+).

Step 3 — Net field along the axis (take direction from −q-q to +q+q as positive):

E=E+−E−=q4πε0[1(r−a)2−1(r+a)2]E = E_+ - E_- = \dfrac{q}{4\pi\varepsilon_0}\left[\dfrac{1}{(r-a)^2} - \dfrac{1}{(r+a)^2}\right].

Step 4 — Combine the fractions:

1(r−a)2−1(r+a)2=(r+a)2−(r−a)2(r2−a2)2=4ar(r2−a2)2\dfrac{1}{(r-a)^2} - \dfrac{1}{(r+a)^2} = \dfrac{(r+a)^2 - (r-a)^2}{(r^2-a^2)^2} = \dfrac{4 a r}{(r^2 - a^2)^2}.

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